The integral I want to solve is $$\int_1^2 \frac {2 \ln(x)}{x+1} dx$$
I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.
Can any one help by giving a solution please?
Your integral evaluates to, $$2(\ln (2)\ln (3) + \operatorname{Li}_2 (-2) + \frac {π^2}{12})$$ Use the identity, $$\int (f.g') dx = f.g - \int (f'.g) dx$$ And the definition of dilogarithm to get your result. Assume $f(x):=\ln x$ and $g'(x):=\frac {1}{1+x}$ and substitute $z:=-x$. $$\operatorname{Li}_2 := \int \frac {\ln (1-t)}{t} dt$$